我怎样才能纠正这个错误,我让我的功能,下面是我的错误我怎样才能解决此警告错误即时得到
Warning: mysql_query() expects parameter 2 to be resource, null given in C:\xampp\htdocs\how are things\15_06_widget_corp-final\includes\functions.php on line 40
Database query failed:
第40行的代码是这样
$subject_set = mysql_query($query, $connection);
这是我的功能码
function get_all_subjects($public = true) {
global $connection;
$query = "SELECT *
FROM subjects ";
if ($public) {
$query .= "WHERE visible = 1 ";
}
$query .= "ORDER BY position ASC";
$subject_set = mysql_query($query, $connection);
confirm_query($subject_set);
return $subject_set;
}
您应该使用'mysqli_query()'而不是'mysql_query()'。对于你的问题,mysql_connect肯定失败了,你给mysql_query的var不是一个有效的资源。检查你的全局'$连接' – MatRt 2013-03-22 05:37:29
看看这是否有助于揭示它> http://stackoverflow.com/questions/5039561/php-sql-database-query-error-message – 2013-03-22 05:39:13
什么是comfirm_query函数? – Nikitas 2013-03-22 05:40:17